A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 *10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium:
-[Cu(CN) ₃²^-]

Question

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 *10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows:

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 *10<sup>-3</sup> M CuNO<sub>3</sub>. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN) <sub>3</sub><sup>2</sup><sup>-</sup>] A) 3.33 M B) 1.67 M C) 2.0 *10<sup>-</sup><sup>3</sup> M D) 1.0 * 10<sup>-</sup><sup>3</sup> M E) none of these

Calculate the following concentrations at equilibrium:
-[Cu(CN) ₃²^-]

Quizplus · Accepted Answer

The Answer of A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 *10 -3 M CuNO 3 . Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN)₃²^-]

A Solution Is Formed by Mixing 50